`2.2 NODE-VOLTAGE AND MESH-CURRENT ANALYSES 77`

`R 11 I 1 − R 12 I 2 − ··· − R 1 NIN = V 1`

−R 21 I 1 + R 22 V 2 − ··· − R 2 NIN = V 2

..

.

`..`

.

−RN 1 I 1 − RN 2 V 2 − ··· + RNNIN = VN

`(2.2.7)`

whereRNNis the sum of all resistances contained in meshN,RJK=RKJis the sum of all

resistances common to both meshesJandK, andVNis the algebraic sum of the source-voltage

rises in meshN, taken in the direction ofIN.

By solving the equations for the unknown mesh currents, other currents and voltages in the

circuit elements can be determined easily.

EXAMPLE 2.2.3

By means of mesh-current analysis, obtain the current in the 10-V source and the voltage across

the 10-resistor in the circuit of Example 2.2.1.

`Solution`

STEP 1:Replace all current sources with shunt resistances by their corresponding Thévenin

equivalents consisting of voltage sources with series resistances. Conductances included in the

circuit are replaced by their equivalent resistances.

In this example, since there are no current sources and conductances, the circuit of Figure

E2.2.1(a) is redrawn as Figure E2.2.3 for convenience.

`+`

`− +`

`−`

`I 1 I 2`

10 V

`20 V`

`5 Ω`

`4 Ω`

`8 Ω 10 Ω`

`20 Ω 25 Ω`

`I 3`

`Figure E2.2.3`

STEP 2:Identify elementary loops (meshes) and choose a mesh-current variable for each

elementary loop, with all loop currents in the same clockwise direction. Mesh currentsI 1 ,I 2 ,

andI 3 are shown in Figure E2.2.3.

STEP 3:In terms of unknown mesh-current variables, write the KVL equations for all meshes

by following the rules for mesh analysis.

`Loop 1 with mesh currentI 1 : ( 5 + 8 + 20 )I 1 − 20 I 2 − 8 I 3 = 10`

Loop 2 with mesh currentI 2 : − 20 I 1 +( 20 + 10 + 25 )I 2 − 10 I 3 = 0

Loop 3 with mesh currentI 3 : − 8 I 1 − 10 I 2 +( 4 + 10 + 8 )I 3 = 20

Rearranging, one gets